Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 11111;
int n;
int x[N], y[N], tot;
vector<pair<int, int> > answerx, answery;
int number[N], pos[N];
int getHigh(int x) {
  int res = -1;
  for (int i = 0; i < 32; i++) {
    if (x >> i & 1) {
      res = i;
    }
  }
  return res;
}
void print(int x[], int i, int j, vector<pair<int, int> > &vec) {
  x[i] ^= x[j];
  vec.push_back(make_pair(i, j));
}
void sswap(int x[], int i, int j, vector<pair<int, int> > &vec) {
  if (i == j) {
    return;
  }
  print(x, i, j, vec);
  print(x, j, i, vec);
  print(x, i, j, vec);
}
void solve(int x[], vector<pair<int, int> > &vec, int &tot) {
  tot = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= tot; j++) {
      if ((x[i] ^ x[j]) < x[i]) {
        print(x, i, j, vec);
      }
    }
    if (x[i] == 0) {
      continue;
    }
    int h = getHigh(x[i]), rank = 0;
    for (int j = 1; j <= tot; j++) {
      if (getHigh(x[j]) > h) {
        rank = j;
      }
    }
    sswap(x, ++tot, i, vec);
    for (int j = tot - 1; j > rank; j--) {
      sswap(x, j, j + 1, vec);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &y[i]);
  }
  int totx, toty;
  solve(x, answerx, totx);
  solve(y, answery, toty);
  if (totx < toty) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= toty; i++) {
    bool find = false;
    for (int j = i; j <= totx; j++) {
      if (getHigh(x[j]) == getHigh(y[i])) {
        find = true;
        sswap(x, i, j, answerx);
      }
    }
    if (!find) {
      printf("-1\n");
      return 0;
    }
    for (int m = 31; m >= 0; m--) {
      if ((x[i] >> m & 1) != (y[i] >> m & 1)) {
        bool find = false;
        for (int j = i + 1; j <= totx; j++) {
          if (getHigh(x[j]) == m) {
            find = true;
            print(x, i, j, answerx);
          }
        }
        if (!find) {
          printf("-1\n");
        }
      }
    }
  }
  totx = toty;
  for (int i = totx + 1; i <= n; i++) {
    print(x, i, i, answerx);
    print(y, i, i, answery);
  }
  for (int i = (int)answery.size() - 1; i >= 0; i--) {
    int a = answery[i].first, b = answery[i].second;
    print(x, a, b, answerx);
  }
  printf("%d\n", (int)answerx.size());
  for (int i = 0; i < (int)answerx.size(); i++) {
    printf("%d %d\n", answerx[i].first, answerx[i].second);
  }
  return 0;
}